回复：回复：1 - (0.1)^(1/6)

Interesting. But let's have a close look. Take 1 minute interval as an example:

If there must be a start in this hour, if the first 59 minute did not see the star, then in the last minute the probability to see the star is 1. You can not use 1- (1-1)^(1/60) here. That means in this case you cannot have consistant probability to see the star in these each minute.

also my understanding is that even though time interval can be very very small, but it cannot be 0, because this is a discrete event scenario.

Thanks.

所有跟帖： 

• 回复：回复：回复：1 - (0.1)^(1/6) -zeal626- ♂ (345 bytes) () 01/30/2011 postreply 23:36:55

• 回复：回复：回复：回复：1 - (0.1)^(1/6) -calligraphy- ♂ (80 bytes) () 01/31/2011 postreply 06:16:37

• 回复：回复：回复：回复：回复：1 - (0.1)^(1/6) -zeal626- ♂ (593 bytes) () 01/31/2011 postreply 23:55:36